AoPS Academy
Prove that
Where 
Prove that
Let 
Prove that
Let 
Prove that
be the unit circle in the complex plane. Let 
be the map given by 
. We define 
and 
for 
. The smallest positive integer 
such that 
is called the period of 
. Determine the total number of points in 
of period 
.
)
such that![\[
x+f(xy)=f(1+f(y))x
\]](http://latex.artofproblemsolving.com/4/3/3/433f95836dad1224754d480aff8432ef033a37cb.png)
for all 
.
. Then we get that: 
,so 
and hence 
, which is a contradiction by 
. Then we get that: 
With the same way as case 
we get that 
and hence 
, which is a contradiction by 
.
. Then we get that: 
, and hence 
, so equivalently we have that 
, which has only 
as a positive integer solution. So, in this case 
,which is the only solution of the given equation.
and 
.
.
.
but, 
which is contradiction.
. then, 
.
.
. obviously contradicition.
.
yields 
is even, so 
. Now if 
we have ![\[2x\geq \nu_{2}(2020^x+2^y)=\min\{2x, 2y_1\}=\nu_{2}(2024^z)=3z.\]](http://latex.artofproblemsolving.com/e/6/3/e631c0799f33d41fc92bc875e0e69e3f038c1ba8.png)
However, 
as 
. Therefore 
, and the equation becomes 
. Modulo 
implies 
is odd, at which point 
. Comparing the 
's of both sides gives 
, so 
, 
for some nonnegative integer 
. Clearly, 
leads to the solution 
. When 
we derive a contradiction from:![\[1>\frac{2020^{3t+1}}{2024^{2t+1}}=\frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)^t\geq \frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)>1.\]](http://latex.artofproblemsolving.com/c/b/2/cb24c7a9ef36eabb403e6fa4982babe2ac46b8b8.png)
, which obviously works. We now prove it is the only one.
.
to get 
. Clearly, we must have 
, so is even and is odd. Let 
, for a positive integer 
. Then, the given equation becomes 
. Note that 
is always odd. Since 
and 
are both even, it follows if 
, then either 
or 
, neither of which are odd. Thus, 
. Because 
, we must have 
, and our claim follows. is .
. Note that when , 
is divisible by all the prime factors of 
, which are 
, , and 
. If 
, by Zsigmondy, there exists some 
that divides 
, and hence it is impossible that 
. Thus, , and we are done. 
, which clearly works.
gives 
. Since 
, if divided either one of or , then we have that one of 
is 
, which is absurd. Hence 
.
are both not multiples of , so they cannot be equal to 
. If 
, then we would have 
, which is absurd, so 
, so 
. Now we have ![\[ 2020^x + 4^x = 2024^z \]](http://latex.artofproblemsolving.com/4/b/e/4be984ca8fd9e9a9078af51ca885348c8cc030f5.png)
If 
, then by Zsigmondy there exists a prime 
not dividing 
that divides 
, absurd. Hence 
must hold.
Working in modulo 
, we obtain:
is 
for all 
.
is 
is 
and 
.
, where 
have the same parity 
.
Working in modulo , we obtain:
is odd number 
is odd number.
.
by 
results:
.
is an even non-negative number and results: the last digit of 
is 
.
must be , hence 
and we obtain
.
in the last relation, results:
is solution of the equation 
.
. by 
results:
.
is odd number; 
and 
are even numbers, hence the last equation has no solutions.
..
, odd number (see 
).
, even number, hence the equation has no solutions.
, even number, hence the equation has no solutions.
.
, where 
(see ).
.
.
.
, contradiction since
(proved many times in the previous posts).
has the unique solution .
. Because of this, we know y must be even and z must be odd. and 
. Prime factorizing the original equation, 
is the smallest factor, then 
, impossible. Similarly, if 2x is the largest factor of 2, 
is impossible. This means 
and the factor of 2 on the LHS is 
since the v2 of 
is at 2. (since its 0 mod 2 and not 0 mod 4). So we have 
. since 
is an integer, set 
Plugging in and dividing by largest factor of 2 we get 
k=0 works, but any larger k dosent work since 
so the only solution is k=0, which represents 
even and 
odd
we get:
if even then: 
which is a contradiction so 
even we get that 
so our equation transforms into the following:
Claim: 
which is a contradiction because we found that 
Claim: 
on both sides and using Lifting the Exponent Lemma (LTE) we get:
Since 
we get that 
Let 
also 
Now by taking we get:
but this isn't true for 
so 
clearly 
so 
is the only solution 
.
is even
gives is odd
is same as 
nonsense
nonsense
again we get is odd
(for a non-negetive integer k)
for 
for 
, so is even and is odd. Now consider of both sides. The RHS is 
, so it's odd. The LHS is 
, and it's odd if and only if . Now write the LHS as 
, and taking mod 4 on the inside sum gives 
. Now we have the relationship 
. We can easily check that 
is a solution, and we see that for , the LHS is strictly greater than the RHS, so there are no solutions 
.
![\[
2020^x + 2^y = 2024^z
\]](http://latex.artofproblemsolving.com/1/e/5/1e5535fd71c16a98e46030aa8eae8404de4804b5.png)
![\[
v_2(2020^x + 2^y) = v_2(2024^z)
\]](http://latex.artofproblemsolving.com/4/0/3/403f86d0ea5bb9977b3bda19223adb3e9c62260f.png)
![\[
v_2(2020) = 2 \Rightarrow v_2(2020^x) = 2x, \quad v_2(2^y) = y, \quad v_2(2024) = 4 \Rightarrow v_2(2024^z) = 4z
\]](http://latex.artofproblemsolving.com/c/2/0/c20aac682f210f5a4fcc284430d5eee48e3bdf1f.png)
![\[
2020^x + 2^y = 2^y(a + 1), \quad \text{where } a \in \mathbb{N}, \text{ odd}
\]](http://latex.artofproblemsolving.com/0/f/2/0f2148ff5c9fa26d0a4fd71a250d11c8771c6e1d.png)
So,![\[
v_2(2020^x + 2^y) = y = 4z \Rightarrow 2x = 4z \Rightarrow x = 2z
\]](http://latex.artofproblemsolving.com/e/b/1/eb16f0372960c1aed9b43478173ad5131e365f96.png)
![\[
2^{2x} + 1 \equiv 0 \pmod{3} \Rightarrow 2^{2x} \equiv -1 \pmod{3}
\]](http://latex.artofproblemsolving.com/c/e/3/ce36c0bd7856b2d5c2c9807b7c408ccbdfc742ac.png)
![\[
2^{2x} \equiv 1 \pmod{3} \Rightarrow \text{Contradiction}
\]](http://latex.artofproblemsolving.com/d/1/7/d1704603ecc3666f2d0498c3f981936d10fd52c9.png)
![\[
2020^x + 2^y = 2^{2x}(a + 1), \quad \text{where } a \text{ is odd}
\Rightarrow v_2 = 2x
\Rightarrow 4z = 2x \Rightarrow z = \frac{x}{2}
\Rightarrow x \text{ must be even}
\]](http://latex.artofproblemsolving.com/c/e/f/cef5778173248f0b5ba255751ce2b51b2cf5bbe0.png)
![\[
2020^x + 2^{2x} = 2024^z
\Rightarrow 4^x(505^x + 1) = 2^{4z}(253^z)
\]](http://latex.artofproblemsolving.com/7/c/9/7c99831ed753bc3fee358d0b7b10c175d0fce540.png)
![\[
2^{2x}(505^x + 1) = 2^{4z}(253^z)
\Rightarrow 505^x + 1 = 2^{2z}(253^z)
\]](http://latex.artofproblemsolving.com/b/9/3/b93b102a69645ef2f1c7e9211a6bb9be85d1a752.png)
![\[
2020^1 + 2^2 = 2020 + 4 = 2024 = 2024^1 \Rightarrow \text{Valid solution}
\]](http://latex.artofproblemsolving.com/6/8/a/68a1da7eb19ad5b3baff0ef83acc0bb5907a636f.png)
![\[
x = 1,\quad y = 2x = 2,\quad 2x + 1 = 3z \Rightarrow z = \frac{2x + 1}{3} = \frac{3}{3} = 1
\]](http://latex.artofproblemsolving.com/7/6/9/769beb4f38b4409b15dd5e6f7a8373d3011f2812.png)
![\[
2020^2 + 2^4 = 4080400 + 16 = 4080416 \\
2024^2 = 4096576 \Rightarrow \text{LHS} < \text{RHS}
\]](http://latex.artofproblemsolving.com/8/a/a/8aa1b5c9ada5421a8ffcf3e48364666b6a29b81e.png)
![\[
2020^3 + 2^6 = 8204080000 + 64 = 8204080064 \\
2024^3 = 8283809024 \Rightarrow \text{LHS} < \text{RHS}
\]](http://latex.artofproblemsolving.com/f/3/6/f360a4f6f8552bf9d485a5f9c658131fa6ccb364.png)
![\[
\boxed{(x, y, z) = (1, 2, 1)}
\]](http://latex.artofproblemsolving.com/2/2/0/220c534c815104162b53ed786d25a370af5ccfdf.png)
that satisfy the equation
we get:
![$2020^x=2024^z-8^z=2016[...]$](http://latex.artofproblemsolving.com/3/8/e/38eaa7c52b664c59165d40a2823dc5b2d0416c86.png)
contradiction since 
but not 
or 
we get that 
and then from inelyalites esily get 
,
and we claim there are no more solutions. Our strategy relies on that the powers of 
is the minimal among the three, then we have a contradiction via mod 
.
and 
.
.
,
.
(as 
for 
and 
for some positive integer 
.
,
,
.
.
is even, but not divisible by 
.
,
.
,
,
and we claim there are no more solutions. Have in mind that 
and 
.
:
and hence, by computations up to 
or similar, 
for some positive integer 
.
(we used 
by Fermat's Little Theorem) and hence, by computations up to 
or similar, 
:
,
.
.
now imply that 
and 
and later by 
)